Posted by Robert on Wednesday, July 25, 2012 at 2:17pm.
This is the last question for a final exam preparation sheet and I can't figure out the answer. Any help would be greatly appreciated. Thank you.
- One very long wire carries current 10.0 A to the left along the x axis. A second very long wire carries current 75.0 A to the right along the line
(y = 0.280 m, z = 0).
(a) Where in the plane of the two wires is the total magnetic field equal to zero? (along the y axis)
(b) A particle with a charge of -2.00 µC is moving with a velocity of 150 Mm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle.
(c) What If? A uniform electric field is applied to allow this particle to pass through this region undeflected. Calculate the required vector electric field.
- Physics - Electromagnetism - Elena, Wednesday, July 25, 2012 at 4:39pm
Magnetic field of the long current carrying wire according to Biot-Savert Law is
The desired point is separated by distance “x” from the 1st current and by distance (d+x) from the 2nd current (along y-axis).
(b) Magnetic force (Lorentz force) is
Since the motion of the particle is along the straight line, then F=0
(c) Electric field has to be applied along the direction of the particle motion (E↑↑ῡ, E↑↓ῡ)
- Physics - Electromagnetism - Robert, Thursday, July 26, 2012 at 9:09am
thank you Elena!
- Physics - Electromagnetism - Robin, Wednesday, June 29, 2016 at 6:10pm
For people of the future, Elena's answer for part B is wrong. It's better to use the equation
F = q*v(cross)B
where B is the sum of the B field created by the 50 A wire and the 30 A wire.
------> v = 150E6 m/s
B_tot = mu/2pi * (50/.18 (-k) + 30/.1 (-k))
= 1.2E-4 (-k) Teslas
Then you take the cross product and eventually get F = -.035j N
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